Competition problem — Number theory

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Number theory Difficulty 5.8 AIME, harder Prove it

We consider the following operation that generates a new number from a given natural number: The given number is represented in an arbitrary integer base $b \geq 2$ , in which it is two-digit with both digits not equal to 0. Then the two digits are swapped, and the result in the digit representation to base $b$ is the new number.

Is it possible, with possibly multiple such operations, to change any number greater than ten to a number less than or equal to ten?

(Theresia Eisenkölbl)

Solution

We show that we can generate a smaller number from any number $>10$ . With this, we will eventually reach a number $\leq 10$ .

If the number $n=2 k+1$ is odd, then we choose the base for the digit representation $b=k$ . Thus, $n=(21)_{k}$ . This gives us the new number $(12)_{k}=k+2$ . Since $k \geq 5$ must hold, the choice of $b=k$ is valid and it follows that $k+2 \leq 2 k-5+25$ are both bases greater than the largest occurring digit and thus valid and it holds that $k+3<2 k$ as desired.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.